Question

Prove that:

sin(π/7)*sin(2π/7)*sin(3π/7)=(√7)/8

Answer 1

Let p=sin(π/7)sin(2π/7)sin(3π/7)
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.

From cos(α + β) and sin(α + β) formulas
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)

Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1):8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.

However, it will take a bit of work to simplify the still messy expression of
P = 2cs³(4c²-1)
= s³(8c³ - 2c) and now apply (1) to get
= s³(4c² + 2c - 1)

Now we take a slight detour:
(2): s²(4c² + 2c - 1)
= (1-c²)(4c² + 2c - 1)
= -4c⁴ -2c³ + 5c² + 2c -1 and now apply (1)
= -4c³ + 3c² + (5/2)c - 1 and apply (1) again to get
= c² + ½c - ½

We now return to our regularly scheduled programming:
P² = (s³(4c² + 2c - 1))² apply (2)
= (s(c² + c/2 - 1/2))²
= s²(c² + ½c - ¼ - ¼)(c² + ½c - ½) apply (2) again
= (¼(c² + ½c - ½) - ¼(1-c²)) * (c² + ½c - ½)
= (4c² + c - 3) * (2c² + c -1) / 16
= (8c⁴ + 6c³ - 9c² - 4c + 3) / 16 apply (1)
= (10c³ - 5c² - 5c + 3) / 16 apply (1) again
= (-5/4 + 3) / 16
= (7/4) / 16

Hence P = √7 /8